How can I know how many real roots this polynomial has?

Solution 1:

Maybe this will work for you

http://www.cs.iastate.edu/~cs577/handouts/polyroots.pdf

Uses Descartes’ rules of sign of the Sturm’s Theorem

Solution 2:

We use the rational roots theorem, which says a root of $$f(x) = x^7 -10x^5+15x+5$$ must be an integer that divides $5$, so immediately we have four possibilities: $\pm 1$ or $\pm 5$. Plugging these into the polynomial we see that $$f(-5) = -46495 \neq 0\\f(-1) = -1 \neq 0 \\f(1) = 11 \neq 0\\f(5) = 46955 \neq 0$$And from this, we see that $f$ has no roots over $\mathbb{Q}$.

Solution 3:

The most direct method that I can currently think of is the following:

Differentiate the polynomial, and make a substitution $w=x^2$. This gives us $$7w^3-50w^2+15,$$ a cubic equation in $w$. There is a general formula for solving for the zeroes of a cubic equations, hence we can find the exact roots of this equation, and in turn, we can find the exact roots of $f'(x)$, which tells us where local max/min occur. Checking the values of the function at these points will tell us how many zeroes the polynomial has.

This is a little bit of a long method and probably not the cleanest; I'll see if I can think of a simpler method, but this is certainly one approach to the problem.